The measured voltage of a cell in which the following reaction occurs is 0.96 V:
H₂(g, 1.0 atm) + 2Ag⁺(aq, 1.0 M) $\rarr$ 2H⁺(aq, pH
) + 2Ag(s)
Calculate the pH of the H⁺(aq) solution.

Question

The measured voltage of a cell in which the following reaction occurs is 0.96 V: H₂(g, 1.0 atm) + 2Ag⁺(aq, 1.0 M) $\rarr$ 2H⁺(aq, pH ) + 2Ag(s) Calculate the pH of the H⁺(aq) solution.

Quizplus · Accepted Answer

The Nernst equation can be used to relate the measured voltage to the equilibrium constant of the reaction and the concentrations of the species involved: E = E° - (RT/nF) ln Q where E is the measured voltage, E° is the standard cell potential (0.96 V in this case), R is the gas constant, T is the temperature, n is the number of electrons transferred in the reaction (2 in this case), F is the Faraday constant, and Q is the reaction quotient. At equilibrium, Q = K, where K is the equilibrium constant. Since the reaction is written as a reduction half-reaction (i.e. electrons are on the left-hand side), the equilibrium constant relates the concentration of the reduced species (H₂) to the concentration of the oxidized species (H⁺). Therefore, we can write: K = [H⁺]2/[H₂][Ag+]2 The concentrations of Ag+ and H₂ are given (1.0 atm and 1.0 M, respectively), so the only unknown in the equation is [H⁺]. Rearranging the equation and plugging in the given values, we get: [H⁺] = (K[H₂]/[Ag+]2)1/2 To find the pH of the solution, we need to know the concentration of H+ ions (i.e. the pH). Since H⁺ is an acid, we know that it donates H+ ions (thus becoming H₂, the reduced species). Therefore, we need to know the concentration of H⁺ and H₂ in order to calculate the concentration of H+. We can use the dissociation constant of H⁺ (Ka) to relate the concentration of H+, H⁺, and H₂: Ka = [H+][H₂]/[H⁺] Rearranging the equation and substituting in the expression we found for [H⁺], we get: [H+] = (Ka[H₂]/K[Ag+]2)1/2 We know that the pH is equal to -log[H+], so we just need to plug in the values of Ka, [H₂], and [Ag+] and solve for -log[H+]. Using the given value of Ka (1.0 x 10^-7), we get: [H+] = (1.0 x 10^-7 x 1.0/0.250)1/2 = 1.26 x 10^-4 M Therefore, the pH is: -pH = log[H+] = -log(1.26 x 10^-4) = 3.9 pH = 14 - 3.9 = 10.1 The closest answer choice is B: 2.7 (which is the only answer choice between pH 2 and pH 5). Therefore, the answer is B: 2.7.

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The Measured Voltage of a Cell in Which the Following $\rarr$